This question is about a conflict between two facts about scalar field theories in 2D. The same sort of question will apply to any scalar field theory with a polynomial potential, but let's specialize to the $\phi^4$ theory. $$\mathcal{L}=\frac{1}{2}\partial\phi\partial\phi+\frac{1}{2}m^2\phi^2+\lambda \phi^4$$ On the one hand there is a spontaneous symmetry breaking phase transition as we vary the potential, and this critical point is described by a single CFT (the same as the Ising model) where $\phi$ has non-trivial scaling dimension.

On the other hand, in perturbation theory most diagrams in the theory are convergent in the UV. The only divergent diagrams involve variations of the loop created by a single propagator beginning and ending at the same point (often referred to as a tadpole). This can be absorbed into a single mass counterterm. The $\lambda$ parameter and the field normalization do not renormalize to any order in perturbation theory.

Concretely what this should mean in terms of a lattice simulation is the following. If we change the lattice spacing, using a finer or coarser grid, we can get the exact same correlation functions as before (at scales much larger than the lattice spacing). All we have to do is change the bare mass parameter in the Lagrangian. The $\lambda$ parameter and the field normalization can stay exactly the same.

Is this non-perturbative statement about correlation functions on the lattice correct? This is a very concrete question which should be able to answered by someone who has done lattice simulations.

I suspect it is not correct because the field $\phi$ somehow needs to pick up an anomalous dimension at the critical point. Also if the parameter $\lambda$ does not renormalize it seems we have a whole family of critical points parametrized by $\lambda$, but it is said that everything is described by the same critical point.

But then if it is not correct, the question is what is causing the field normalization and $\lambda$ to have cutoff dependence? To be clear, I understand that perturbation theory does not work at the critical point, but these things do not renormalize to any order in perturbation theory, and they should start renormalizing well before we hit the massless point.

(To see a version of me asking the exact same question a slightly different way, look at the first edit)

  • $\begingroup$ Like divergent integrals, finite integrals are cutoff dependent as well. The coupling $\lambda$ would run (renormalize) regardless whether its loop correction is finite (super-renormalizable) or divergent (renormalizable). So what's is wrong with being cutoff dependent? $\endgroup$
    – MadMax
    Oct 28, 2019 at 21:29
  • $\begingroup$ @MadMax, But if it is finite, that means all the cutoff dependent parts vanish as you take the cutoff to infinity. So if you take the cutoff to be much larger than the scale you are interested in, there is essentially no cutoff dependence. So I don't understand why you say the coupling would run regardless. True, there is a non-trivial relationship between the bare $\lambda$ parameter and some kind physical scattering amplitude, but once we fix that for one cutoff, it continues to hold for all other cutoffs. $\endgroup$
    – octonion
    Oct 29, 2019 at 5:30
  • $\begingroup$ There is a subtle difference between the renormalization scale (which dictates the running of coupling constants) and the cutoff scale. In the Wilsonian renormalization group scheme, unfortunately, these two scales are mixed together. The renormalization scale (hence running) is always there, regardless whether it's a finite or divergent renormalization. $\endgroup$
    – MadMax
    Oct 29, 2019 at 14:30
  • $\begingroup$ If you can distinguish between the 5 scales here physics.stackexchange.com/questions/498977/…, you would have a rather complete picture. $\endgroup$
    – MadMax
    Oct 29, 2019 at 14:39
  • $\begingroup$ @MadMax, In terms of bare quantities, all of my correlation functions depend on bare $\lambda_0$, the bare mass $m_0$, and the cutoff $\Lambda$. I can get rid of explicit cutoff dependence by rewriting my formulas in terms of a renormalized mass $m(\Lambda,\lambda_0,m_0)$, say the correlation length. I don't have to do anything with $\lambda_0$ or the normalization of the fields. In the Callan-Symanzik equation for these correlation functions there should be no terms associated to beta functions of $\lambda_0$ or the anomalous dimension. (cont.) $\endgroup$
    – octonion
    Oct 29, 2019 at 22:12

2 Answers 2


I'm going to attempt a different answer, based on what I think you're trying to get at in the comments, especially since you've been asking how things look when one simulates field theory on a lattice.

So let's set up our theory with an explicit cutoff, $\Lambda > 0$, which can be a placeholder for anything you wish (an inverse lattice spacing, a hard momentum cutoff, etc.). We'll just take Euclidean $\phi^4$ theory: $$ \mathcal{Z} = \int_{\Lambda} \mathcal{D} \phi \exp\left[ - \int d^2 x \left( \frac{1}{2} \left( \nabla \phi \right)^2 + \frac{m^2}{2} \phi^2 + \frac{\lambda}{4!} \phi^4 \right) \right]. $$ In what follows, we will never completely drop the bare parameters $m^2$, $\lambda > 0$, but I will introduce particular observables and renormalized couplings. This way, we are being very careful in taking the $\Lambda \rightarrow \infty$ limit (which formally causes the bare couplings to diverge), which is helpful because this limit is secretly the scaling limit. The scaling limit everywhere away from the critical point is a free Gaussian, where the fields then have trivial scaling dimensions. I think this last fact is what may be causing your confusion.

First, to answer your question:

Concretely what this should mean in terms of a lattice simulation is the following. If we change the lattice spacing, using a finer or coarser grid, we can get the exact same correlation functions as before (at scales much larger than the lattice spacing). All we have to do is change the bare mass parameter in the Lagrangian. The 𝜆 λ parameter and the field normalization can stay exactly the same.

Is this non-perturbative statement about correlation functions on the lattice correct? This is a very concrete question which should be able to answered by someone who has done lattice simulations.

The answer to the question is no: for a particular fine-tuned value of $m_c^2$, whose exact value depends on your choice of $\lambda$ and $\Lambda$, this theory "flows" to the Ising CFT in the scaling limit. You will likely be interested in the following reference which studies this theory numerically: https://journals.aps.org/prd/abstract/10.1103/PhysRevD.91.085011. (I admit I haven't read it since it came out in 2015 and I was a much weaker physicist.) The approach to the scaling limit can be thought of in two different ways:

  1. Take $\Lambda$ fixed, but consider studying our theory on very long length scales. If we compute the two-point function at the critical point, we find $$ \lim_{x \rightarrow \infty} \Gamma^{(2)}(x)^{-1} = \lim_{x \rightarrow \infty} \langle \phi(x) \phi(0) \rangle \propto \frac{1}{|\Lambda x|^{1/4}}. $$ with some dimensionless non-universal coefficient. If we deviate the coupling $m^2$ a small amount from the critical point, we find that this correlator asymptotically decays as $e^{-|x|/\xi}$, where $\xi \propto (m^2 - m_c^2)^{-\nu}$ where $\nu = 1$ in our case.
  2. Consider a family of theories which take varying values of $m^2$, $\lambda$, and $\Lambda$. Now compute exactly the observables like $\Gamma^{(n)}$ and $\xi$ as analytic functions of $m^2$, $\lambda$, and $\Lambda$ (easier said than done, I know). So you have access to the exact functions $$ \Gamma^{(n)}(x) = \Gamma^{(n)}(x,m^2,\lambda,\Lambda), \qquad \xi = \xi(m^2,\lambda,\Lambda). $$ Now it turns out that there is a precise limit of the form $\Lambda \rightarrow \infty$, $m^2 \rightarrow m_c^2$ ($m_c$ is non-universal, and also generically divergent in this process), such that $\xi \rightarrow \infty$ and the $\Gamma^{(n)}$'s take the functional form of the Ising CFT (with appropriate overall factors of $\Lambda$ for dimensional analysis). A rather large complication here involves finding the right way to take $m^2 \rightarrow m_c^2$: for a critical theory with critical exponent $\nu$, one needs the dependence $$ \xi \sim (m^2 - m_c^2)^{-\nu}. $$ The usual procedure for dealing with this is to systematically (within perturbation theory) eliminate the bare quantities $m^2$, $\lambda$, in terms of interesting "observable" quantities like $\xi$ or $\lambda_R(k) = \Gamma^{(4)}(k)$, and then the limit of $\Lambda \rightarrow \infty$ which leaves $\xi$ in the scaling regime is less opaque. This whole procedure is way messier conceptually than the first, but it's actually conducive to analytic calculations.

These two pictures are equivalent because in both procedures you're taking the ratio of all large length scales ($\xi$ and $x$) and your one small length scale ($\Lambda^{-1}$) to infinity.

This is not in conflict with perturbation theory. Let's now choose $\Lambda$ to be a hard cutoff in momentum space, so all momenta satisfy $k^2 \equiv |\mathbf{k}|^2 \leq \Lambda^2$. Now we can choose to compute the correlation length, which is given by the inverse two-point function evaluated at $k^2 = 0$. I find $$ \xi^{-2} \equiv \Gamma^{(2)}(k = 0) = m^2 + \frac{\lambda}{8 \pi} \log \left[ 1 + \frac{\Lambda^2}{m^2} \right] + \frac{\lambda^2}{64 \pi^2 m^2 (m^2 + \Lambda^2)}\log \left[ 1 + \frac{\Lambda^2}{m^2} \right] + \frac{\lambda^2}{m^4} F(m^2/\Lambda^2) + \cdots. \tag{*}\label{*} $$ Here I have neglected to explicitly compute the finite sunset diagram, represented here by $F(m^2/\Lambda^2)$, but the terms shown will suffice to show the main point.

In a QFT course, you now introduce a renormalized mass, $$ m_R^2 = m^2 - \frac{\lambda}{8 \pi} \log \left[ 1 + \frac{\Lambda^2}{m^2} \right], $$ and now the theory is completely finite in the limit $\Lambda \rightarrow \infty$; explicitly $$ \xi^{-2} = m_R^2 + \frac{\lambda^3}{m_R^4} F(0). $$ But this is only part of the scaling limit! One also wants to take $\xi \rightarrow \infty$, which involves taking $m_R \rightarrow 0$ at this order in perturbation theory. Returning back to Eq. \eqref{*}, one sees that in perturbation theory with a finite cutoff, non-analytic contributions to $\xi$ occur in both the $\Lambda \rightarrow \infty$ and $m^2 \rightarrow 0$ limits, and while we can deal with the dependence on $\Lambda$ to all orders with renormalization, we are not taking into account the non-analytic behavior due to the $m^2 \rightarrow m_c^2$ limit which appears at every order. This is unacceptable, because as outlined above, correctly obtaining the dependence of physical observables on $(m^2 - m_c^2)$ is crucial for taking the scaling limit.

The way that the famous Wilson-Fisher $\epsilon = 4 - d$ expansion gets around this is that the IR and UV divergences are both logarithmic just under the upper-critical dimension, resulting in terms like $\log(m/\Lambda)$. Then the Callan-Symanzik equations can be used to resum the logarithms into bonafide power laws (the coefficients of the logs become related to critical exponents). But in the $d=2$ case, one either works with a massive theory, where the early $\Lambda \rightarrow \infty$ limit shoots us deep into the adjacent phases, or one tries to work with the massless theory, which is plagued by incurable IR divergences.

  • $\begingroup$ Thanks, this is helpful. I think I somehow was imagining it was an attracting fixed point, so the dimension would smoothly go from 0 to 1/4 as we go to the IR limit, but of course it is not and it is massive. I see you are saying in your last paragraphs that the anomalous dimension is coming from the IR behavior after all. I thought that might be the case with your last answer, but again I was somehow thinking it was well defined to talk about the anomalous dimension away from the critical point. I'll think a bit and probably award you best answer. $\endgroup$
    – octonion
    Oct 31, 2019 at 5:06
  • $\begingroup$ @octonion I recently came across the following paper which looks at a similar question: Amit et al 1980 J. Phys. A: Math. Gen. 13 585. It studies the 2D sine-Gordon model, which is superrenormalizable in the same sense as $\phi^4$ theory. High energy physicists had shown that the renormalized theory ceased to exist past a critical coupling, but Kosterlitz showed that the critical coupling describes the KT transition. The resolution in this paper is similar to what I say at the end of my answer - one needs the analytic dependence of an infinite number of individually finite diagrams. $\endgroup$ Nov 26, 2019 at 20:42
  • $\begingroup$ Thanks, I'll take a look at it sometime. I've heard the idea that the critical coupling in 2D sine-Gordon is connected with the KT transition before, but I've always been meaning to look at that in more detail too. $\endgroup$
    – octonion
    Nov 26, 2019 at 20:58
  • $\begingroup$ Right, the sine-Gordon model is dual to the XY model. There's a nice "derivation" in the first part of these lecture notes, which then show the RG: qpt.physics.harvard.edu/phys268b/Lec2_XYmodel_duality.pdf . I found the Amit paper while trying to resolve the KT picture with that in high-energy physics, as given in Sidney Coleman's work on sine-Gordon-Thirring duality for example. $\endgroup$ Nov 27, 2019 at 1:19

The problem is the distinction between a theory being "perturbatively super-renormalizable" (or renormalizable/non-renormalizable), and the non-perturbative renormalization of scaling dimensions which you can see in the "exact" theory. The issue appears here because $\phi^4$ theory is only perturbative in $\lambda$ away from the massless point. For example, if you calculate the leading correction to the inverse propagator of $\phi$ perturbatively, you will find $$ G(p = 0)^{-1} = m^2 \left[ 1 - c_1 \frac{u}{m^2} \right], $$ signaling that the leading perturbative correction to the tree-level result becomes larger than the leading-order expression, so perturbation theory is breaking down. If we considered the finite-$p$ dependence of certain correlations functions, we would similarly find terms which are IR-divergent. For example, at $m^2 = 0$ the four-point function would look like $$ \Gamma^{(4)}(p) = u + c_2 \frac{u^2}{p^2} \log(p/\mu), $$ and the small-$p$ behavior is clearly non-perturbative. This is sort of the mirror case of the bad UV behavior in non-renormalizable theories.

So you cannot actually calculate any scaling dimensions at the critical point using perturbation theory - your intuition obtained from the perturbative renormalization constants can only be applied away from the critical point, where the theory flows to a Gaussian theory with zero anomalous dimension.

  • $\begingroup$ I understand there are IR problems in perturbation theory, and perturbation theory should not work at the critical point, but I don't think this answers any of my questions. I rewrote my question to be clearer. Thanks for your response $\endgroup$
    – octonion
    Oct 28, 2019 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.